Ncert solutions

Ncert solutions

Grade 8

Squares and Square Roots | Exercise 6.3

Question 9

Q10) Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

L.C.M. of 8, 15 and 20 is 120.

Prime factors of 120 = 2 x 2 x 2 x 3 x 5

Here, prime factors 2, 3 and 5 have no pair. Therefore 120 must be multiplied by 2 x 3 x 5 to make it a perfect square.

\therefore120\times2\times3\times5=3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.

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subject-cta

Question 9

Q10) Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

Solution:

L.C.M. of 8, 15 and 20 is 120.

Prime factors of 120 = 2 x 2 x 2 x 3 x 5

Here, prime factors 2, 3 and 5 have no pair. Therefore 120 must be multiplied by 2 x 3 x 5 to make it a perfect square.

\therefore120\times2\times3\times5=3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.

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Book a free class now

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subject-cta
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